Lecture Signal processing: Transform analysis of LTI systems - Nguyễn Công Phương

Nguyễn Công Phương

SIGNAL PROCESSING
Transform Analysis of LTI Systems


Contents
I. Introduction
II. Discrete – Time Signals and Systems
III. The z – Transform
IV. Fourier Representation of Signals
V. Transform Analysis of LTI Systems
VI. Sampling of Continuous – Time Signals
VII.The Discrete Fourier Transform
VIII.Structures for Discrete – Time Systems
IX. Design of FIR Filters
X. Design of IIR Filters
XI. Random Signal Processing
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2


Transform Analysis
of LTI Systems
1.
2.
3.
4.
5.
6.

7.
8.
9.
10.
11.

Sinusoidal Response of LTI Systems
Response of LTI Systems in the Frequency Domain
Distortion of Signals Passing through LTI Systems
Ideal and Practical Filters
Frequency Response for Rational System Functions
Dependency of Frequency Response on Poles and Zeros
Design of Simple Filters by Pole – Zero Placement
Relationship between Magnitude and Phase Responses
Allpass Systems
Invertibility and Minimum – Phase Systems
Transform Analysis of Continuous – Time Systems
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3


Sinusoidal Response of LTI
Systems (1)
x[ n ] = z → y[n ] = H ( z ) z , all n;
H ( z ) = ∑ h[ k ]z
∞

n

n

−k

k =−∞

z = e jω → x[n] = e jωn → y[n] = H (e jω )e jωn , all n
jω

H (e ) = H ( z ) z =e jω =
jω

jω

H (e ) = H (e ) e
x[n ] = Ae

j ( ω n +φ )

j∠H ( e jω )

∞

− jωk
h
[
k
]
e
∑

k =−∞

= H R (e jω ) + jH I (e jω )
jω

→ y[n] = A H (e ) e

j [ωn +φ +∠H ( e jω )]

The response of a stable LTI system to a complex exponential sequence
is a complex exponential sequence with the same frequency,
only the amplitude and phase are changed by the system.
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4


Sinusoidal Response of LTI
Systems (2)
Ax jφx jωn Ax − jφx − jωn
x[n ] = Ax cos(ωn + φx ) =
e e + e e
2
2

x[n ] = Ae

j ( ω n +φ )

jω

→ y[n] = A H (e ) e

j [ωn +φ +∠H ( e jω )]

Ax jφx jωn
Ax
jφx j [ ωn +∠H ( e jω )]
jω
x1[n ] =
e e → y1[n ] =
H (e ) e e
2
2
Ax − jφx − jωn
Ax
− jφ x j [ − ωn +∠H ( e − jω )]
− jω
x2 [n] =
e e
→ y2 [ n ] =
H (e ) e e
2
2
Ax
Ax
j [ω n +φ x +∠H ( e jω )]
j [ − ωn −φx +∠H ( e − jω )]
jω

− jω
→ y[ n ] =
H (e ) e
+
H (e ) e
2
2
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5


Sinusoidal Response of LTI
Systems (3)
Ax
Ax
j[ ωn +φ x +∠H ( e jω )]
j[ − ωn −φ x +∠H ( e − jω )]
jω
− jω
y[ n ] =
H (e ) e
+
H (e ) e
2
2

H (e − jω ) = H (e jω ) ;

∠H (e− jω ) = −∠H (e− jω )

Ax
Ax
j [ ωn +φx +∠H ( e jω )]
− j [ ωn +φx +∠H ( e jω )]
jω
jω
→ y[ n ] =
H (e ) e
+
H (e ) e
2
2
Ax
j [ ωn +φx +∠H ( e jω )]
− j [ ωn +φ x +∠H ( e jω )]
jω
=
H (e ) {e
+e
}
2
= Ax H (e jω ) cos ωn + φx + ∠H (e jω )
x[ n ] = Ax cos(ωn + φx ) → y[n ] = Ay cos(ωn + φ y )
Ay = Ax H ( e jω )

( magnitude response/gain )

φ y = ∠H (e jω ) + φx (phase response)
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6


Ex. 1

Sinusoidal Response of LTI
Systems (4)

Find the frequency response of the system y[n] = ay[n – 1] + bx[n],

–1 < a < 1.

x[n] = e jωn → y[n] = H (e jω )e jωn
y[n − 1] = H (e jω )e jω ( n −1)

→ H ( e jω )e jωn = aH ( e jω )e jω ( n −1) + be jωn
b
y[ n] = ay[ n − 1] + bx[ n] → H (e jω ) =
1 − ae − jω

e − jω = cos ω − j sin ω
→ H (e jω ) = b

1 − cos ω
a sin ω
−
jb
1 − 2a cos ω + a 2
1 − 2a cos ω + a 2


b
jω
 H (e ) =
2

1
−
2
a
cos
ω
+
a
→
∠H (e jω ) = atan −a sin ω

1 − a cos ω
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7


Sinusoidal Response of LTI
Systems (5)

Ex. 1

Find the frequency response of the system y[n] = ay[n – 1] + bx[n],

b

H (e jω ) =

1 − 2a cos ω + a 2

∠H (e jω ) = atan

;

–1 < a < 1.

−a sin ω
1 − a cos ω

Magnitude

2

1.5

1

0.5

-3

-2

-1

0
omega (rad)

1

2

3

-3

-2

-1

0
omega (rad)

1

2

3

0.6
0.4

Phase

0.2
0
-0.2
-0.4

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8


Sinusoidal Response of LTI
Systems (6)
x[n] = e jωn → y[n ] = H (e jω )e jωn , all n
x[n] = e

jω n

n

n

k =0

k =0

u[n] → y[n ] = ∑ h[k ]x[n − k ] = ∑ h[k ]e jω ( n −k )
 ∞
 ∞
− jω k  jω n

− jωk  jωn
=  ∑ h[k ]e
e
−
h
[
k
]
e

∑
e
 k =0

 k =n +1

jω

= H (e )e

jω n

y steady state [ n ]

 ∞
− jωk  jωn
−  ∑ h[k ]e
e
 k = n +1


ytransient response [ n ]

ytr [n] ≤

∞

∑

k = n +1

∞

h[k ] ≤ ∑ h[k ] < ∞
k =0

lim y[n] = H (e jω )e jωn = yss [n]
n →∞

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9


Ex. 2

Sinusoidal Response of LTI
Systems (7)

Given a system h[n] = 0.7nu[n] and an input x[n] = cos(0.03πn)u[n], find the output?
y[ n] = h[ n]* x[n ] → Y ( z ) = H ( z ) X ( z )

H ( z) =

1
,
1 − 0.7 z −1

z > 0.7

1 − cos(0.03π ) z −1
X ( z) =
,
−1
−2
1 − 2 cos(0.03π ) z + z

z >1

1 − cos(0.03π ) z −1
→ Y ( z) =
(1 − 0.7 z −1 )[1 − 2 cos(0.03π ) z −1 + z −2 ]

1 − 2 cos(0.03π )z −1 + z −2 = 0
∆ = 4 cos2 α − 4 = − 4 sin 2 α

−1

→ ( z )1,2

2 cos α ± −4 sin 2 a
=

= cos α ± j sin α
2

= e ± jα = e ± j 0.03π

→ 1 − 2 cos(0.03π ) z −1 + z −2 = ( z −1 − e j 0.03π )( z −1 − e − j 0.03π ) = (1 − e j 0.03π z −1 )(1 − e − j 0.03π z −1 )
1 − cos(0.03π ) z −1
→ Y ( z) =
(1 − 0.7 z −1 )(1 − e j 0.03π z −1 )(1 − e − j 0.03π z −1 )
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10


Ex. 2

Sinusoidal Response of LTI
Systems (8)

Given a system h[n] = 0.7nu[n] and an input x[n] = cos(0.03πn)u[n], find the output?
1 − cos(0.03π ) z −1
K1
K2
K3
Y ( z) =
=
+
+
(1 − 0.7 z −1 )(1 − e j 0.03π z −1 )(1 − e − j 0.03π z −1 ) 1 − 0.7 z −1 1 − e j 0.03π z −1 1 − e − j 0.03π z −1

→ 1 − cos(0.03π ) z −1 = K1 (1 − e j 0.03π z −1 )(1 − e − j 0.03π z −1 ) + K 2 (1 − 0.7 z −1 )(1 − e − j 0.03π z −1 ) +
+ K3 (1 − 0.7 z −1 )(1 − e j 0.03π z −1 )
z −1 = 0.7 −1 → 1 − cos(0.03π )0.7 −1 = K1 (1 − e j 0.03π 0.7 −1 )(1 − e − j 0.03π 0.7 −1 ) +
+ K 2 (1 − 0.7 × 0.7 −1 )(1 − e − j 0.03π 0.7 −1 ) + K 3 (1 − 0.7 × 0.7 −1 )(1 − e j 0.03π 0.7 −1 )
1 − 0.7 −1 cos(0.03π )
→ K1 =
(1 − 0.7 −1 e j 0.03π )(1 − 2 e − j 0.03π )

1 − 0.7−1 cos(0.03π )
1 − 0.7−1 cos(0.03π )
=
=
−1 j 0.03π
− 1 − j 0.03π
−2
1 − 0.7 e
− 0 .7 e
+ 0 .7
1 + 0.7 −2 − 0.7 −1( e j 0.03π + e − j 0.03π )
e jθ = cos θ + j sin θ
e − jθ = cos θ − j sin θ

→ e j 0.03π + e− j 0.03π = 2 cos(0.03π )
1 − 0.7 −1 cos(0.03π )
→ K1 =
= −2.1504
1 + 0.7 −2 − 0.7 −12 cos(0.03π )

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11


Ex. 2

Sinusoidal Response of LTI
Systems (9)

Given a system h[n] = 0.7nu[n] and an input x[n] = cos(0.03πn)u[n], find the output?
1 − cos(0.02π ) z −1
−2.1504
K2
K3
Y ( z) =
=
+
+
(1 − 0.5z −1 )(1 − e j 0.02π z −1 )(1 − e − j 0.02π z −1 ) 1 − 0.7 z −1 1 − e j 0.03π z −1 1 − e − j 0.03π z −1

→ 1 − cos(0.03π ) z −1 = −2.1504(1 − e j 0.03π z −1 )(1 − e − j 0.03π z −1 ) + K 2 (1 − 0.7 z −1 )(1 − e − j 0.03π z −1 ) +
+ K 3 (1 − 0.7 z −1 )(1 − e j 0.03π z −1 )
z −1 = e − j 0.03π → 1 − cos(0.03π )e − j 0.03π = K 2 (1 − 0.7e − j 0.03π )(1 − e − j 0.03π e − j 0.03π )

1 − e − j 0.03π cos(0.03π )
→ K2 =
= 1.6120∠ − 0.2140
(1 − 0.7 e − j 0.03π )(1 − e − j 2×0.03π )
z −1 = e j 0.03π → 1 − cos(0.03π )e j 0.03π = K 3 (1 − 0.7e j 0.03π )(1 − e j 0.03π e j 0.03π )

1 − e − j 0.03π cos(0.03π )

→ K3 =
= 1.6120∠0.2140
(1 − 0.7e j 0.03π )(1 − e j 2×0.03π )
→ Y ( z) =

−2.1504 1.6120∠ − 0.2140 1.6120∠0.2140
+
+
1 − 0.7 z −1
1 − e j 0.03π z −1
1 − e − j 0.03π z −1
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12


Ex. 2

Sinusoidal Response of LTI
Systems (10)

Given a system h[n] = 0.7nu[n] and an input x[n] = cos(0.03πn)u[n], find the output?
Y ( z) =

−2.1504 1.6120∠ − 0.2140 1.6120∠0.2140
+
+
1 − 0.7 z −1
1 − e j 0.03π z −1
1 − e − j 0.03π z −1

X ( z) =

K
n
→
x
[
n
]
=
Kp
u[n ]
−1
1 − pz

→ y[n ] = −2.1504(0.7)n u[ n] + (1.6120∠ − 0.2140)(e j 0.03π ) n u[ n] + (1.6120∠0.2140)( e − j 0.03π )n u[n ]

= [ −2.1504(0.7)n + 1.6120e − j 0.2140 ( e j 0.03π ) n + 1.6120e j 0.2140 (e j 0.03π ) n ]u[n ]
= {−2.1504(0.7)n + 1.6120[ e j ( 0.03π n−0.2140 ) + e −( 0.03π n −0.2140) ]}u[ n]
= −2.1504(0.7) n u[n] + 1.6120 × 2 cos(0.03π n − 0.2140)u[ n]
ytransient response [ n ]

ysteady state [ n ]

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13


Sinusoidal Response of LTI

Systems (11)

Ex. 2

Given a system h[n] = 0.7nu[n] and an input x[n] = cos(0.03πn)u[n], find the output?
y[ n] = −2.1504(0.7)n u[ n] + 1.6120 × 2 cos(0.03π n − 0.2140)u[ n]
ytransient response [ n ]

y steady state [ n ]

4
x[n]
y[n]
3

2

1

0

-1

-2

-3

-4

0

20

40

60

80
Time index (n)

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100

120

140

160

14


Transform Analysis
of LTI Systems
1.
2.

Sinusoidal Response of LTI Systems
Response of LTI Systems in the Frequency Domain

a)
b)
c)

3.
4.
5.
6.
7.
8.
9.
10.
11.

Response to Periodic Inputs
Response to Aperiodic Inputs
Power Gain

Distortion of Signals Passing through LTI Systems
Ideal and Practical Filters
Frequency Response for Rational System Functions
Dependency of Frequency Response on Poles and Zeros
Design of Simple Filters by Pole – Zero Placement
Relationship between Magnitude and Phase Responses
Allpass Systems
Invertibility and Minimum – Phase Systems
Transform Analysis of Continuous – Time Systems
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15

Response to Periodic Inputs (1)
N −1

x[n ] = ∑ ck( x )e

x[ n] = x[ n + N ],

j

2π
kn
N

k =0

2π
kn
N

j

e
N −1

x[n] = ∑ c e
k =0

( x)

k

c
c

= H (e

Py =

1
N

N −1

∑
n=0

2π
kn
N

j

N −1

2π
k
N

j

2π
k
N

(x)
k

) c

2π
k
N

→ ∑ c H (e
k =0

= H (e

(y)
k

(y)
k

j

→ y[n ] = H (e

j

(x)
k

N −1

2π
k
N

2π
kn
N

)e

∠c

( y)
k

= ∠H ( e

N −1

y[n] = ∑ ck( y ) = ∑ H (e
2

k =0

j

2π
kn
N

= y[n]

−∞< k < ∞

)ck( x ) ,

,

j

)e

j

2

k =0

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j

2π
k

N

j

2π
k
N

) + ∠ck( x )
2

) ck( x )

2

16


Response to Periodic Inputs (2)
Ex.
1,
0,

Given a periodic sequence x[ n] = 

0≤n< N
N ≤n
with fundamental period M

and a system function h[n] = anu[n]. Find the output sequence?

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17


Transform Analysis
of LTI Systems
1.
2.

Sinusoidal Response of LTI Systems
Response of LTI Systems in the Frequency Domain
a)
b)
c)

3.
4.
5.
6.
7.
8.
9.
10.
11.

Response to Periodic Inputs
Response to Aperiodic Inputs

Power Gain

Distortion of Signals Passing through LTI Systems
Ideal and Practical Filters
Frequency Response for Rational System Functions
Dependency of Frequency Response on Poles and Zeros
Design of Simple Filters by Pole – Zero Placement
Relationship between Magnitude and Phase Responses
Allpass Systems
Invertibility and Minimum – Phase Systems
Transform Analysis of Continuous – Time Systems
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18


Response to Aperiodic Inputs
Y (e jω ) = H (e jω ) X ( e jω )

 Y (e jω ) = H (e jω ) X (e jω )
→
jω
jω
jω
∠Y (e ) = ∠H (e ) + ∠X (e )

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19

Power Gain
jω

Gain in dB = H (e )

jω

dB

= 10 log10 H (e )

Y (e jω ) = H (e jω ) X (e jω ) → Y (e jω )

dB

= H (e jω )

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dB

2

+ X (e jω )

dB

20

Transform Analysis
of LTI Systems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Sinusoidal Response of LTI Systems
Response of LTI Systems in the Frequency Domain
Distortion of Signals Passing through LTI Systems
Ideal and Practical Filters
Frequency Response for Rational System Functions
Dependency of Frequency Response on Poles and Zeros
Design of Simple Filters by Pole – Zero Placement
Relationship between Magnitude and Phase Responses
Allpass Systems
Invertibility and Minimum – Phase Systems
Transform Analysis of Continuous – Time Systems
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21

Distortion of Signals Passing
through LTI Systems (1)
A system has distortionless response if the input signal x[n]
and the output signal y[n] have the same “shape”

y[n ] = Gx[ n − nd ],

G >0

→ Y (e jω ) = Ge − jωnd X (e jω )
Y ( e jω )
− jωnd
→ H (e ) =
=
Ge
X ( e jω )
jω

 H (e jω ) = G
→ 
jω
∠H (e ) = −ωnd
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22


Distortion of Signals Passing
through LTI Systems (2)

y[n ] = Ax H (e jω ) cos ωn + φx + ∠H (e jω )
jω


φ
∠
H
(
e
)
jω
x
= Ax H (e ) cos ω n + +

ω
ω

 

τ phase delay (ω ) = −

∠H (e jω )

ω

d Ψ(ω )
τ group delay (ω ) = −
dω
x[n ] = s[n]cos ωc n → y[n ] ≈ H (e jωc ) s[n − τ gd (ωc )]cos{ωc [n − τ pd (ωc )]}
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23


Transform Analysis
of LTI Systems
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.

Sinusoidal Response of LTI Systems
Response of LTI Systems in the Frequency Domain
Distortion of Signals Passing through LTI Systems
Ideal and Practical Filters
Frequency Response for Rational System Functions
Dependency of Frequency Response on Poles and Zeros
Design of Simple Filters by Pole – Zero Placement
Relationship between Magnitude and Phase Responses
Allpass Systems
Invertibility and Minimum – Phase Systems
Transform Analysis of Continuous – Time Systems
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24


Ideal and Practical Filters (1)
H lp ( e jω )

1

1

−π

−ωc 0 ωc

π

ω

−π −ωc

Lowpass

Highpass

Bandpass

Bandstop

H bp ( e jω )

0

ωl

ωu

π

ω

−π
−ωu

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ωc π

0

1

1
−π
−ωu −ωl

H hp ( e jω )

ω

H bs (e jω )

−ωl 0 ωl

ωu

π

ω
25